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Proof of logarithmic properties.

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    There are two log properties that I have to prove:
    1) Explain why ln(b1/n)=(1/n)ln(b) for b>0, set b=an

    2) Explain why ln(ar)=rln(a) for any r in Q and a>0, ie r is rational.

    2. Relevant equations


    3. The attempt at a solution

    In a previous question I have already proved that ln(an)=nln(a), where n is a natural number. What I'm unsure about, is how is this any different? For 1), I'm not sure why you set b=an? Wouldn't you get ln((an)1/n) = ln(a)? I'm not sure how this helps me find the solution.

    Similarly for 2), I'm unsure how it is any different to proving that ln(an)=nln(a).

    Any help would be great!
  2. jcsd
  3. Sep 29, 2010 #2


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    The obvious difference is that you proved a result [tex] m ln(a)=ln(a^m)[/tex] when m is a natural number, and now they want you to do it when m is not a natural number. The proof that you used for this will not work when m is not a natural number (unless you did something really clever) because at some point you assume that it was, probably in order to write am as a*a*a...*a a multiplied together m times (which doesn't make sense if m=1/2 for example)

    If b=an, what does ln(b) and ln(b1/n) become? Use it to do something cool
  4. Oct 2, 2010 #3
    Thanks for your help. I worked out part 1, but I'm still unsure on part 2. For proving the law for a natural number I did:
    Let y=ln(a), for a>0,
    ey=a, and if we raise each side to the power n we get:
    y*n = ln(an), but y = ln(a)
    n*ln(a) = ln(an)

    So I'm still unsure how it's any different for any rational number?
    Any help would be great :)
  5. Oct 2, 2010 #4


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    Use the definition of Q, namely that r =m/n, where m,n are in N.
  6. Oct 2, 2010 #5
    Ah I get it now! Thanks :)
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